CPhill

The area of the cross-section  will be   BC  * CG

Note....BC  = AD    

So...the area is

BC * CG  =

AD * CG =

6 * 12  =

72 cm^2

The width of the rectangle is easy...it  = 8 mm

If we connect the center of each circle, we have an equilateral triangle with sides of 2

And the height of this triangle can be found thusly :

tan 60  = height / 2

√3  =  height/ 2

2√3 mm = height

So....the height of the rectangle  = 2√3 + 2 + 2  =  [2√3 + 4] mm

So...the area  is      width of rectangle * height  of rectangle =

8  ( 2√3 + 4 ) mm^2  =

16√3 + 32 mm^2

The volume of the full tank  is  51 * 39 * 28   =  55692 cm^3 

And 95%  of this is 52907.4 cm^3

So.....the difference between these  is the amount of water required left to fill the tank  = 2784.6 cm^3  (1)

The volume of the log is    (3.14) (5)^2 (44)   = 3454 cm^3   (2)

So....the volume of water that spills out of the tank is just  (2) - (1)  =

3454  - 2784.6  =  

669.4  cm^3

6)The slant height and radius of a cone are 4 and 1, respectively. Unrolling the curved surface gives a circular sector with center angle n degrees. Find n.

If the radius of the cone is 1, the perimeter of the circular sector  is  2pi (radius) =

2pi (1)  =    2pi

So........when we unroll the curved surface, we will have a circular sector with a perimeter of 2pi  and a radius  of 4

So....we can find the central angle [in radians] thusly :

2pi  = radius * theta in radians

2pi  = 4 * theta in radians      divide both sides by 4

pi/2  = theta in radians

And.....pi/2 in radians =  90°

5)Find the volume of a cone that has 5 as its base radius and a lateral surface area of 65pi.

The lateral surface area, A,  is given by   

A  = pi * radius * slant height

65 pi  = pi * 5 * slant height       divide both sides by 5 pi

13  = slant height

We can use the Pythagorean Theorem to find the height of the cone

height  =  √ [ slant height^2  - radius^2 ]  = √[13^2 - 5^2]  = √ [144]  = 12

So....the volume of the cone is

(1/3) pi (radius^2) (height )  =  (1/3) pi * 5^2 * 12  =  

(1/3) pi * 300  =

100 pi  units^3

4)The volume of a cylinder is 100. Another cylinder has twice the height but half the base radius. What is the volume of the second cylinder?

100  = pi (r^2) h

Solving for the height, we have

h  = 100/ (pi *r^2)

So....twice the height  =  200 / (pi * r^2)

So....the volume, V, of the second cylinder  can be expressed as :

V  =  pi (r/2)^2* (200) / (pi * r^2)  =  (pi/pi) (r^2/r^2) (200/ 4)   =  (1)(1)*50  = 50

Note....at   x  = 0, y   = 3

So...we are looking for the next x value to the right where y = 3, again

And....when x  = pi/2   , y is at 3, again

So...the period  is from  the "top of the curve" to the "top of the curve"  =   pi/2

This one is a little difficult......

Here's the form of the function that we need :

f(x)  = Asin (Bx)  + C

The amplitude, A, can be determined by  taking the [high point - the midline point ]  =

8.3  - 4.35  =  3.95

"C"   will be the midline point  = 4.3

B"  is the harder to determine.....we want to divide 2pi by 24  = pi / 12  = B

So  our function is

f(x)  =  3.95sin [ pi/12 * x ]  + 4.35

Here's the graph :

https://www.desmos.com/calculator/1ox0tmscf7

Note that  at  =  0 [ midnight ] we are at 4.35 ft

At x = 6  [ 6AM]  we are at high tide = 8.3 ft

At   x = 12  [noon] we are back to 4.35 ft

At x  = 18  [ 6PM] we are at low tide  = 0.4 ft

At x  = 24  [ midnight, again ] we are back to the midline

The first statement is true

The second statement is false.....at 0 seconds, the loading height is 34 ft.....but the lowest point on the wheel is 4 ft

The third statement is also true.....the  radius =  [ highest point - lowest point ]/ 2  =

[64  - 4 ] / 2  =   60/ 2  =    30 ft

The last statement is not true....at 0 seconds, the height is 34 ft, at 2.5 sec, the height is 64 ft, at 5 sec, the height is back at 34 ft, at 7.5 sec the height is at 4ft and at 10sec, we are back where we started

So......the wheel makes one rotation in 10 seconds

So....statements   1  and 3 are the only ones that are true

f(x)  =  3sin (x)

The "3" out front just vertically stretches the graph of the sine by a factor of 3

Here's the regular sine graph and the given function  on the same axis :

https://www.desmos.com/calculator/oiwp1jl38e